# Prove that $\sum\limits_{cyc}\frac{a}{\sqrt{a+3b}}\geq\sqrt{a+b+c+d}$

Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that: $$\frac{a}{\sqrt{a+3b}}+\frac{b}{\sqrt{b+3c}}+\frac{c}{\sqrt{c+3d}}+\frac{d}{\sqrt{d+3a}}\geq\sqrt{a+b+c+d}$$ I tried Holder, AM-GM and more, but without success.

• I have an advice, set $a+3b=s^2,b+3c=t^2.\cdots$ – Isaac YIU Math Studio 23 hours ago
• @Isaac YIU Math Studio Let $a+3b=x^2$, $b+3c=y^2$, $c+3d=z^2$ and $d+3a=t^2$, where $x$, $y$, $z$ and $t$ are positives. Thus, $a=\frac{-x^2+3y^2-9z^2+27t^2}{80}$ and it's nothing. – Michael Rozenberg 23 hours ago
• I think it may have help – Isaac YIU Math Studio 22 hours ago
• I don't see it. :( – Michael Rozenberg 21 hours ago
• Not nothing! It made a better expansion of the left side – Isaac YIU Math Studio 21 hours ago

Hello Michael Rozenberg . I start my proof with a classical substitution .

Put: $A=a$; $AB=b$; $AC=c$; $AD=d$ We get : $$\frac{1}{\sqrt{1+3B}}+\frac{B}{\sqrt{B+3C}}+\frac{C}{\sqrt{C+3D}}+\frac{D}{\sqrt{D+3}}\geq \sqrt{1+B+C+D}$$

Make a new substitution like this :

$x=B$; $C=xh$; $D=x\beta$

We get : $$\frac{1}{\sqrt{1+3x}}+\frac{\sqrt{x}}{\sqrt{1+3h}}+\frac{\sqrt{x}h}{\sqrt{h+3\beta}}+\frac{x\beta}{\sqrt{x\beta+3}}\geq \sqrt{1+x(1+h+\beta)}$$ We can rewriting the inequality like this : $$\frac{1}{\sqrt{1+3x}}+\frac{\sqrt{x}}{\sqrt{1+3h}}+\frac{\sqrt{x}h}{\sqrt{h+3\beta}}+\sqrt{x\beta+3}+\frac{-3}{\sqrt{x\beta+3}}-\sqrt{1+x(1+h+\beta)}\geq 0$$

Now we introduce the following function :

$$f(x)=\frac{1}{\sqrt{1+3x}}+\frac{\sqrt{x}}{\sqrt{1+3h}}+\frac{\sqrt{x}h}{\sqrt{h+3\beta}}+\sqrt{x\beta+3}+\frac{-3}{\sqrt{x\beta+3}}-\sqrt{1+x(1+h+\beta)}$$

The idea is to integrate the function $f$ we get :

$$F(x)=\frac{2}{3}\sqrt{1+3x}+\frac{2}{3}(x)^{\frac{3}{2}}\frac{1}{\sqrt{1+3h}}+\frac{2}{3}(x)^{\frac{3}{2}}\frac{h}{\sqrt{h+3\beta}}+\frac{2}{3\beta}(x\beta+3)^{\frac{3}{2}}-\frac{6}{\beta}\sqrt{x\beta+3}-\frac{2}{3(1+h+\beta)}(1+x(1+h+\beta))^{\frac{3}{2}}$$

Now the idea is to note that the function $f$ is positiv if the function $F$ is increasing . Namely when we have : $F(X)\leq F(Y)$ with $X\leq Y$

Here we take $X=x-\epsilon$ and $Y=y+\epsilon$

We cut the function $F$ in others to a question of space

We have :

$g(x-\epsilon)-g(x+\epsilon)=\frac{2}{3}\sqrt{1+3(x-\epsilon)}-\frac{2}{3}\sqrt{1+3(x+\epsilon)}$

$u(x-\epsilon)-u(x+\epsilon)=\frac{2}{3}(x-\epsilon)^{\frac{3}{2}}\frac{1}{\sqrt{1+3h}}+\frac{2}{3}(x-\epsilon)^{\frac{3}{2}}\frac{h}{\sqrt{h+3\beta}}-(\frac{2}{3}(x+\epsilon)^{\frac{3}{2}}\frac{1}{\sqrt{1+3h}}+\frac{2}{3}(x+\epsilon)^{\frac{3}{2}}\frac{h}{\sqrt{h+3\beta}})$

$v(x-\epsilon)-v(x+\epsilon)=\frac{2}{3\beta}((x-\epsilon)\beta+3)^{\frac{3}{2}}-\frac{2}{3\beta}((x+\epsilon)\beta+3)^{\frac{3}{2}}$

$q(x+\epsilon)-q(x-\epsilon)=\frac{6}{\beta}\sqrt{(x+\epsilon)\beta+3}+\frac{2}{3(1+h+\beta)}(1+(x+\epsilon)(1+h+\beta))^{\frac{3}{2}} -(\frac{6}{\beta}\sqrt{(x-\epsilon)\beta+3}+\frac{2}{3(1+h+\beta)}(1+(x-\epsilon)(1+h+\beta))^{\frac{3}{2}} )$

Now we have the following elementary inequalities with $n\geq x^6$:

$g(x-\epsilon)-g(x+\epsilon)\leq -\frac{2}{3}\frac{\epsilon}{n} x^6$

$u(x-\epsilon)-u(x+\epsilon)\leq -\frac{2\epsilon}{3n}x^6(\frac{1}{\sqrt{1+3h}}+\frac{h}{\sqrt{h+3\beta}})$

$v(x-\epsilon)-v(x+\epsilon)\leq -\frac{2\epsilon}{3n}x^6$

$q(x+\epsilon)-q(x-\epsilon)\leq \frac{\epsilon}{n}(\frac{2}{3})x^6$

So you have just to add the elementary inequalities to get the result (and derivate) If you have questions tell me .

• I think there are mistakes in your integration. – Michael Rozenberg Mar 26 '17 at 18:25
• I am going to correct .Thanks! – max8128 Mar 27 '17 at 8:21

WLOG, assume that $$d = \min(a,b,c,d)$$.

Squaring both sides and using AM-GM, it suffices to prove that \begin{align} &\sum_{\mathrm{cyc}} \frac{a^2}{a+3b} + \sum_{\mathrm{cyc}} \frac{4ab}{a+3b + b+3c} + \frac{4ac}{a+3b + c+3d} + \frac{4bd}{b+3c + d+3a} \\ \ge\ & a + b + c + d. \end{align} After clearing the denominators, it suffices to prove that $$f(a, b, c, d)\ge 0$$ where $$f(a,b,c,d)$$ is a homogeneous polynomial of degree $$11$$.

The Buffalo Way works. Let $$c=d+s, \ b = d+t, \ a = d+r; \ s, t, r\ge 0$$. We have $$f(d+r, d+t, d+s, d) = q_9d^9 + q_8d^8 + \cdots + q_1d + q_0.$$ Here $$q_9 = 12582912r^2+16777216rs-20971520rt+12582912s^2-20971520st+12582912t^2$$, etc.

It suffices to prove that $$q_9, q_8, \cdots, q_0 \ge 0$$. I used Mathematica Resolve and the Buffalo Way to verify it. We are done. However, nice or simple proof for $$q_9, q_8, \cdots, q_0 \ge 0$$ is expected.